An iron sphere of mass $20\times {10}^{-3}\mathrm{kg}$ falls through a viscous liquid with terminal velocity 0.5 m/sec. The terminal velocity (in m/se) of another iron sphere of mass $54\times {10}^{-2}\mathrm{kg}$ is

Let d and r be the density and radius of iron, respectively.

$\frac{4}{3}{\mathrm{\pi r}}^3\times \mathrm{d}=20\times {10}^{-3}\mathrm{\_\_\_\_\_\_}\left(1\right)$

$0.5\mathrm{m}/\mathrm{s}=\frac{2{\mathrm{gr}}^2(\mathrm{T}-\mathrm{\sigma })}{9\mathrm{\eta }}\mathrm{\_\_\_\_\_\_\_}\left(2\right)$

in the second case,

$\frac{4}{3}{{\mathrm{\pi r}}_1}^3\times \mathrm{d}=540\times {10}^{-3}$

$\frac{{{\mathrm{r}}_1}^3}{{{\mathrm{r}}_2}^3}=27$

${\mathrm{r}}_1=3\mathrm{r}$

Now the terminal velocity,

${\mathrm{T}}_{\mathrm{V}}=\frac{2\mathrm{g}{\left(3\mathrm{r}\right)}^2(\mathrm{T}-\mathrm{\sigma })}{9\mathrm{\eta }}$

${\mathrm{T}}_{\mathrm{V}}=\frac{2{\mathrm{gr}}^2(\mathrm{T}-\mathrm{\sigma })}{9\mathrm{\eta }}\times 9=9\times 0.5\mathrm{m}/\mathrm{s}=4.5\mathrm{m}/\mathrm{s}$

Hence the correct answer is 4.5.